Two surfaces in a 4-manifold whose algebraic intersection number is zero Suppose $X$ is a smooth closed oriented 4-manifold, and $\Sigma_1,\Sigma_2$ are smoothly embedded compact oriented surfaces in $X$. Suppose they intersect transversally at two points with different sings, so that their algebraic intersection is zero: $[\Sigma_1]\cdot [\Sigma_2]=0$. Then can we take other representatives $\Sigma_1', \Sigma_2'$ with $[\Sigma_i']=[\Sigma_i]\in H_2(X;\Bbb Z)$ so that $\Sigma_1'\cap \Sigma_2'$ is empty?
Here is a special case. Let $X=\Bbb CP^2\sharp \Bbb CP^2$ with $H_2(X;\Bbb Z)=H_2(\Bbb CP^2;\Bbb Z)\oplus H_2(\Bbb CP^2;\Bbb Z)=\Bbb Z\oplus \Bbb Z$. Consider the classes $(1,1)$ and $(1,-1)$. Their algebraic intersection is zero. Can we take disjoint surfaces representing these classes?
 A: Yes, this can be done by tubing one surface along the other.
Suppose that you have two intersection points $p_+, p_- \in \Sigma_1 \cap \Sigma_2$ of opposite signs. Suppose also that $\Sigma_1$ and $\Sigma_2$ are connected (which we can assume, without loss of generality).
Choose a path $\gamma \subset \Sigma_1$ connecting $p_+$ to $p_-$ and avoiding all other intersection points with $\Sigma_2$. The restriction to $\gamma$ of the normal bundle to $\Sigma_1$ is trivial, and we can choose a cylinder $C = [-1,1] \times D^2$ embedded in $X$ such that $C \cap \Sigma_2$ is two disc neighbourhoods $D_\pm$ of $p_\pm$, which comprise the top and bottom of the cylinder, $\{\pm1\}\times D^2 \subset C$. Call $S$ the sides of the cylinder, i.e. $[-1,1]\times S^1$.
Now replace $\Sigma_2$ with $\Sigma_2' = \Sigma_2 \setminus (D_+ \cup D_-) \cup S$ (and secretly smooth corners along the way). This removes two points of intersection from $\Sigma_2 \cap \Sigma_1$, and keeps the same homology classes. ($C$ itself gives a 3-chain whose boundary is $\Sigma'_2 - \Sigma_2$.
Note that this operation increases the genus. If you wanted to do this while keeping the genus, you can't always do it. This is related to the minimal genus problem in 4-manifolds.
You can do it if, for instance, you find enough Whitney discs (allowing you to perform the so-called Whitney trick).
